Question

In double slit experiment, the distance between two slits is $0.6mm$ and these are illuminated with light of wavelength $4800\mathring{A}$. The width of dark fringe on the screen at a distance $120cm$ from slits will be:

• A. $8\times {10}^{-4}$ radian
• B. $60\times {10}^{-4}$ radian
• C. $6\times {10}^{-4}$ radian
• D. $16\times {10}^{-4}$ radian

Answer 1

The correct option is A $8\times {10}^{-4}$ radian

In Young's double slit experiment, angular width $\left(\theta \right)$ of the fringe is given by,

$\theta =\frac{\lambda }{d}$

where, $\lambda$=wavelength of light used $=4800\mathring{A}=4800\times {10}^{-10}m$

$d$=distance between 2 slits$=0.6mm=0.6\times {10}^{-3}m$

Now,

$\theta =\frac{4800\times {10}^{-10}}{0.6\times {10}^{-3}}=8\times {10}^{-4}radian$